The discussion revolves around calculating the displacement of a car that travels 10 km due east and then 12 km northwest, focusing on the application of vector principles in physics.
Participants are actively engaging with the problem, with some offering insights into the mathematical principles involved. There is an acknowledgment of mistakes and a request for clarification on how to determine the angle needed for calculations.
Some participants express uncertainty about the angle between the two displacement vectors, which is crucial for applying the cosine rule. There is a mention of needing to draw a diagram to aid understanding.
I got 8.6. We can't tell where your mistake is until you post your working.Said90 said:A car travels 10 km due EAST then 12. Km NW what is Magnitude of the car's displacement.
The answer that I got in my book is 8.8
Could anyone-please-explain ?The Attempt at a Solution
6.6 km[/B]
haruspex said:I got 8.6. We can't tell where your mistake is until you post your working.
It is not a right-angled triangle. Do you know a rule for finding the third side of a triangle given two sides and the angle between them?Said90 said:***I made a mistake .. The answer in the book is 8.6***I've just applied Pythagorean theorem to get the answer !
144 = 100+X^2
Could please explain your answer?
haruspex said:It is not a right-angled triangle. Do you know a rule for finding the third side of a triangle given two sides and the angle between them?
Draw yourself a diagram.Said90 said:Yea..I got the idea
C^2=a^2+b^2- 2bc COSa
But still don't know how to get the angle's value